Well. the 1050 ohm source in parrallel with the 5000 ohm load puts the
resistance seen by the tank at around 867 ohms.

The circuit reactances would need to be RL/Q = 174 ohms.

I think you multiplied instead of divided. The calculations above assume
that the Q of the elements that you use is large enough not to impact the
calculations. You can calculate the equivilent parrallel resistance of the
inductor you select, and add it in parrallel with the 867 ohms to assess the
impact of a finite-Q inductor.

Don